Find all digits $a,b,c,x,y,z$ for which $\overline{abc},\overline{xyz}$ and $\overline{abcxyz}$ are all perfect squares (no leading zeros allowed).
Proposed by Adrian Andreescu, Dallas, Texas
Solution:
Let $\overline{abcxyz}=n^2$, where $n \in \mathbb{N}$. Since $\overline{abc}$ must be a perfect square, it's easy to see that $n \in \{317,348,349,380,412,443,475,506,538,570,601,633,696,728,759,791,854,886,949\}$. An easy check shows that the given conditions are satisfied if and only if $n \in \{380,475,570\}$. So, $$(a,b,c,x,y,z) \in \{(1,4,4,0,0,0),(2,2,5,6,2,5),(3,2,4,9,0,0)\}.$$
Solution:
Let $\overline{abcxyz}=n^2$, where $n \in \mathbb{N}$. Since $\overline{abc}$ must be a perfect square, it's easy to see that $n \in \{317,348,349,380,412,443,475,506,538,570,601,633,696,728,759,791,854,886,949\}$. An easy check shows that the given conditions are satisfied if and only if $n \in \{380,475,570\}$. So, $$(a,b,c,x,y,z) \in \{(1,4,4,0,0,0),(2,2,5,6,2,5),(3,2,4,9,0,0)\}.$$
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